Theorem frss 2921

Description: Subset theorem for the founded predicate. Exercise 1 of [TakeutiZaring] p. 31.

Assertion

Ref	Expression
frss	|- (A (_ B -> (R Fr B -> R Fr A))

Proof of Theorem frss

Step	Hyp	Ref	Expression
1		sstr2 2071	. . . . . 6 |- (x (_ A -> (A (_ B -> x (_ B))
2	1	com12 11	. . . . 5 |- (A (_ B -> (x (_ A -> x (_ B))
3	2	anim1d 560	. . . 4 |- (A (_ B -> ((x (_ A /\ x =/= (/)) -> (x (_ B /\ x =/= (/))))
4	3	imim1d 28	. . 3 |- (A (_ B -> (((x (_ B /\ x =/= (/)) -> E.y e. x (x i^i {z | zRy}) = (/)) -> ((x (_ A /\ x =/= (/)) -> E.y e. x (x i^i {z | zRy}) = (/))))
5	4	19.20dv 1289	. 2 |- (A (_ B -> (A.x((x (_ B /\ x =/= (/)) -> E.y e. x (x i^i {z | zRy}) = (/)) -> A.x((x (_ A /\ x =/= (/)) -> E.y e. x (x i^i {z | zRy}) = (/))))
6		dffr2 2919	. 2 |- (R Fr B <-> A.x((x (_ B /\ x =/= (/)) -> E.y e. x (x i^i {z | zRy}) = (/)))
7		dffr2 2919	. 2 |- (R Fr A <-> A.x((x (_ A /\ x =/= (/)) -> E.y e. x (x i^i {z | zRy}) = (/)))
8	5, 6, 7	3imtr4g 553	1 |- (A (_ B -> (R Fr B -> R Fr A))

Syntax hints:   -> wi 3   /\ wa 223  A.wal 954   = wceq 956  {cab 1463   =/= wne 1585  E.wrex 1646   i^i cin 2046   (_ wss 2047  (/)c0 2280   class class class wbr 2619   Fr wfr 2915

This theorem is referenced by:  freq2 2923  wess 2936

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-7 962  ax-gen 963  ax-8 964  ax-10 966  ax-12 968  ax-17 971  ax-4 973  ax-5o 975  ax-6o 978  ax-9o 1123  ax-10o 1140  ax-16 1210  ax-11o 1218  ax-ext 1459

This theorem depends on definitions:  df-bi 147  df-or 224  df-an 225  df-ex 981  df-sb 1172  df-clab 1464  df-cleq 1469  df-clel 1472  df-ral 1649  df-rex 1650  df-v 1812  df-dif 2049  df-un 2050  df-in 2051  df-ss 2053  df-nul 2281  df-sn 2412  df-pr 2413  df-op 2416  df-br 2620  df-fr 2917

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